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Solving differential equation

  1. Dec 21, 2005 #1
    i'm trying to find a mirror shape which focuses a light at some specific point [itex]x_0[/itex]
    the initial equation i derived for determining the shape of the mirror is:
    (assuming that light rays fall parallel to x axis - light source is very far from the mirror)
    f(x) is the shape i'm trying to determine

    [tex]
    x_0=-\frac{f(x)-\tan(2\arctan(\frac{df}{dx}))x}{\tan(2\arctan(\frac{df}{dx}))}
    [/tex]

    basicly this is an expression for a line passing through point [itex]x_0[/itex] and point on
    f(x) where light reflected.
    so [itex]\tan(2\arctan(\frac{df}{dx}))[/itex] is a incline of this line


    from the initial equation i got to this point and i'm not sure what to do next::confused:

    [tex]
    \frac{f(x)}{x-x_0}=\tan(2\arctan(\frac{df}{dx}))
    [/tex]
     
    Last edited: Dec 21, 2005
  2. jcsd
  3. Dec 21, 2005 #2
    ok, i took the formula
    [tex]
    \tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}
    [/tex]

    using this formula i can simplify the equation to become:

    [tex]
    \frac{2\frac{df}{dx}}{1-{(\frac{df}{dx})}^2}=\frac{f(x)}{x-x_0}
    [/tex]

    is it possible to solve it?
     
  4. Dec 21, 2005 #3
    i get [itex] f = \pm \sqrt{ (x+c)^2 - (x-x_0)^2 } [/itex] as
    a solution. (where c is an arbitrary constant, hopefully
    not the same as -x0.)
     
  5. Dec 21, 2005 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, I'm impressed with what you have done so far!
    I assume you mean the light is focussed at the point (x0,0).

    You might as well multiply out the terms in the square root:
    [tex]f(x)= \sqrt{x^2+ 2cx+ c^2- x^2+2x_0x- x_0^2}[/tex]

    You can't determine c from the information given: any parabola with focus at (x0,0) will focus light as required.
    And don't write [itex]f(x)= \pm \sqrt{...}[/itex]. A function is single valued. You can, of course, write x as a function of y.
     
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